Open subset of R written as a countable union of pairwise disjoint open intervals?

[dumbQuestion]

I wasn't sure if I should post this in the analysis or topology forum, but this seems to be closely related to compactness so I thought I'd post it here. When dealing with ℝ, the following theorem seems to be really important:"Every non-empty open set G in ℝ can be uniquely expressed as a finite or countably infinite union of pairwise disjoint open intervals in ℝ"Unfortunately, I have a very difficult time figuring out this proof even though apparently it seems like it's supposed to be pretty obvious. Sadly, the theorem doesn't even make intuitive sense to me. The proofs I see all start at like this:

(I'm just copying from this example document http://www.math.louisville.edu/~lee/RealAnalysis/IntroRealAnal-ch05.pdf , the proof on the top of page 5-7, because this seems to be a common way to tackle this proof)

"Let G be open in R. For x ∈ G let α_x = inf {y | (y, x] ⊂ G} and β_x =
sup {y| [x, y) ⊂ G}. The fact that G is open implies α_x < x < β_x. Define I_x = (α_x, β_x). Then I_x ⊂ G"

Here is where my confusion begins: isn't I_x G itself? We know G is an open interval on ℝ, so G = (a,b) for some a < b. So isn't α_x = a and β_x =
sup {y| [x, y) ⊂ G} = b? I mean just by def of sup and inf? I feel this is where all my confusion stems from and if I could clear up my faulty logic at this step I could digest the remainder of the proof.

 

[jgens]

Not all open sets are intervals in $\mathbb{R}$. Take the set $(-1,0) \cup (0,1)$ for example.

 

[micromass]

Why do you think G is an open interval?? The theorem merely says that G is an open set. Not all open sets are open intervals! An easy example is $(0,1)\cup (2,3)$.

 

[dumbQuestion]

aaaaah, thanks guys! I'm facepalming at the moment...

 

[blue_raver22]

I can understand your confusion with this theorem. It may seem counterintuitive at first, but it is actually a fundamental result in topology. Let me try to explain it in a different way that might help you understand it better.

First, let's define what an open subset of ℝ is. An open subset of ℝ is a set G such that for any point x in G, there exists an open interval (α, β) containing x that is completely contained within G. In other words, every point in G has a neighborhood that is also contained in G.

Now, let's consider an open subset G of ℝ. For any point x in G, we can define αx and βx as the infimum and supremum, respectively, of all the numbers y such that (y, x] and [x, y) are both contained in G. This means that αx is the largest number that is less than x and is still in G, and βx is the smallest number that is greater than x and is still in G.

Next, we define Ix as the open interval (αx, βx). This interval is also contained in G, because αx and βx are the largest and smallest numbers in G that are less than x and greater than x, respectively.

Now, let's consider all the points x in G and the corresponding intervals Ix. We can see that these intervals are pairwise disjoint, meaning that no two of them overlap. This is because if two intervals Ix and Iy overlap, then there would be a point z in both intervals, but this is impossible since z would have to be both less than x and greater than y, which is a contradiction.

Finally, we can express G as a union of all these intervals Ix. This is because every point in G is also in one of the intervals Ix, and no two intervals overlap. So we can say that G is a countable union of pairwise disjoint open intervals.

I hope this explanation helps you understand the theorem better. It may still seem counterintuitive, but it is a result that has been proven and used in many areas of mathematics. As a scientist, it is important to accept and understand results like this, even if they may seem confusing at first.